= Solution
The <mass shell> constraint and the conserved <comoving momentum> give
$$
\frac{dt}{ds}=\sqrt{1+\frac{(p^C)^2}{m^2a^2}},\qquad \frac{d\mathbf x}{ds}=\frac{\mathbf p^C}{ma^2}.
$$
Dividing yields
$$
\boxed{\frac{d\mathbf x}{dt}=\frac{\mathbf p^C}{a\sqrt{(p^C)^2+m^2a^2}}.}
$$
The comoving direction is constant because $\mathbf p^C$ is constant. With $a(t_0)=1$ and physical <momentum> magnitude $p_0>0$ at $t_0$, we have $p^C=p_0$. Integrating the speed along that fixed direction gives
$$
\boxed{d=\int_0^{t_0}\frac{dt}{a(t)}\frac1{\sqrt{1+[ma(t)/p_0]^2}}.}
$$
This is a distance travelled, rather than a light-ray distance: the extra factor is the particle's proper peculiar speed $v=p^K/\sqrt{m^2+(p^K)^2}$. For a particle at rest, $p_0=0$, the distance is zero, understood separately or as the limit. The formula assumes collisionless motion over the interval. Its existence at the lower endpoint depends on the early <scale factor>; in <radiation domination> the integral is finite.
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